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Wednesday, 12 July 2006

ac.commutative algebra - Locally square implies square

OK, I've got it. There is no such local criterion for squareness.



Let k be a field of characteristic not 2. Take the ring of triples (f,g,h)ink[t]3, subject to the conditions that f(1)=g(1), g(1)=h(1) and h(1)=f(1). Consider the element (t2,t2,t2). If this were a square, its square root would have to be (pmt,pmt,pmt). But two of those pm's would be the same sign, and t evaluated at 1 and at 1 are not equal.



Now, to check that (t2,t2,t2) is everywhere locally a square. Geometrically, we are talking about three lines glued into a triangle. Any prime ideal has a neighborhood which is contained in the union of two neighboring lines, say the first two. On the first two lines, (t,t,1) is a square root of (t2,t2,t2).



For the suspicious, an algebraic proof. Set u1=(0,(1+t)/2,(1t)/2) and let u2 and u3 be the cyclic permutations thereof. We have u1+u2+u3=1 so, in any local ring, one of the ui must be a unit. WLOG, suppose that u1 is a unit. Notice that u1(1,t,t)2=u1(t2,t2,t2). So, in a local ring where u1 is a unit, (1,t,t) is a square root of (t2,t2,t2).

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